Y=6x+20 substitution
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Answer:
Step-by-step explanation:
Hello!
X: length of human pregnancies from conception to birth.
X~N(μ;σ²)
μ= 264 day
σ= 16 day
If the variable of interest has a normal distribution, it's the sample mean, that it is also a variable on its own, has a normal distribution with parameters:
X[bar] ~N(μ;σ²/n)
When calculating a probability of a value of "X" happening it corresponds to use the standard normal: Z= (X[bar]-μ)/σ
When calculating the probability of the sample mean taking a given value, the variance is divided by the sample size. The standard normal distribution to use is Z= (X[bar]-μ)/(σ/√n)
a. You need to calculate the probability that the sample mean will be less than 260 for a random sample of 15 women.
P(X[bar]<260)= P(Z<(260-264)/(16/√15))= P(Z<-0.97)= 0.16602
b. P(X[bar]>b)= 0.05
You need to find the value of X[bar] that has above it 5% of the distribution and 95% below.
P(X[bar]≤b)= 0.95
P(Z≤(b-μ)/(σ/√n))= 0.95
The value of Z that accumulates 0.95 of probability is Z= 1.648
Now we reverse the standardization to reach the value of pregnancy length:
1.648= (b-264)/(16/√15)
1.648*(16/√15)= b-264
b= [1.648*(16/√15)]+264
b= 270.81 days
c. Now the sample taken is of 7 women and you need to calculate the probability of the sample mean of the length of pregnancy lies between 1800 and 1900 days.
Symbolically:
P(1800≤X[bar]≤1900) = P(X[bar]≤1900) - P(X[bar]≤1800)
P(Z≤(1900-264)/(16/√7)) - P(Z≤(1800-264)/(16/√7))
P(Z≤270.53) - P(Z≤253.99)= 1 - 1 = 0
d. P(X[bar]>270)= 0.1151
P(Z>(270-264)/(16/√n))= 0.1151
P(Z≤(270-264)/(16/√n))= 1 - 0.1151
P(Z≤6/(16/√n))= 0.8849
With the information of the cumulated probability you can reach the value of Z and clear the sample size needed:
P(Z≤1.200)= 0.8849
n= 10.24 ≅ 11 pregnant women.
I hope it helps!